I ran cipher to wipe free space on an HDD and it completed 3 passes in around 40 minutes, while sdelete with 1 pass needed 6 hours. Is there a reason for the time difference, or are both of them doing something different?
I used sdelete -c C:Theoretically, cipher /w:C (where C is the letter of the drive or partition you want to wipe free space on) is equivalent to sdelete -c C: /accepteula which means 3 passes over free space
However the documentation for earlier versions than v1.6 reversed the functions of -c and -z --especially confusing because when running with either, it always starts by displaying "zeroing"...
-z Zero free space--fills with zeroes only so single pass
-c Clean free space to military DoD standards--fills with random data which is obviously no good for TRIM, by first writing zeroes (0x00), then 255s (0xFF) and finally with random numbers so writes three times
Also, SDelete v2.0 had a severe bug making it take 10-28x as long to do the job, because it stalled at 100% doing nothing. I don't know about the current version v2.04 as I have always been happy with the performance of v1.61. So it could be some combination of the above or v2.04 just still takes longer than versions before v2.
I tried the current version and 3 releases prior.Just for the record: What version of Recuva?